Transcript Lecture 3 Notes

MECh300H Introduction to Finite
Element Methods
Finite Element Analysis (F.E.A.) of 1-D
Problems
Historical Background
• Hrenikoff, 1941 – “frame work method”
• Courant, 1943 – “piecewise polynomial
interpolation”
• Turner, 1956 – derived stiffness matrice for truss,
beam, etc
• Clough, 1960 – coined the term “finite element”
Key Ideas: - frame work method
piecewise polynomial approximation
Axially Loaded Bar
Review:
Stress:
Stress:
Strain:
Strain:
Deformation:
Deformation:
Axially Loaded Bar
Review:
Stress:
Strain:
Deformation:
Axially Loaded Bar – Governing
Equations and Boundary
Conditions
• Differential Equation
d 
du 
EA
(
x
)
 f ( x)  0


dx 
dx 
0 x L
• Boundary Condition Types
• prescribed displacement (essential BC)
• prescribed force/derivative of displacement
(natural BC)
Axially Loaded Bar –Boundary
Conditions
• Examples
• fixed end
• simple support
• free end
Potential Energy
• Elastic Potential Energy (PE)
- Spring case
Unstretched spring
PE  0
Stretched bar
1 2
PE  kx
2
x
- Axially loaded bar
undeformed:
deformed:
- Elastic body
PE 
PE  0
L
1
PE    Adx
20
1 T
σ εdv

2V
Potential Energy
• Work Potential (WE)
f
P
B
A
L
WP    u  fdx  P  uB
f: distributed force over a line
P: point force
u: displacement
0
• Total Potential Energy
L
L
1
   Adx   u  fdx  P  uB
20
0
• Principle of Minimum Potential Energy
For conservative systems, of all the kinematically admissible displacement fields,
those corresponding to equilibrium extremize the total potential energy. If the
extremum condition is a minimum, the equilibrium state is stable.
Potential Energy + Rayleigh-Ritz
Approach
Example:
f
P
A
B
Step 1: assume a displacement field u   aii x  i  1 to n
i
 is shape function / basis function
n is the order of approximation
Step 2: calculate total potential energy
Potential Energy + Rayleigh-Ritz
Approach
Example:
f
P
A
B
Step 3:select ai so that the total potential energy is minimum
Galerkin’s Method
Example:
f
P
A
d 
du 
EA
(
x
)
 f ( x)  0


dx 
dx 
u  x  0  0
EA( x )
du
P
dx x  L
B
Seek an approximation u~ so
du~ 
d 

w
EA
(
x
)

f
(
x
)

dV  0
V i  dx 
dx 

u~  x  0   0
du~
EA( x )
P
dx x  L
In the Galerkin’s method, the weight function is chosen to be the same as the shape
function.
Galerkin’s Method
Example:
f
P
A
du~ 
d 

w
EA
(
x
)

f
(
x
)

dV  0
V i  dx 
dx 

1
2
3
B
L
L
du~ dwi
du~
  EA( x )
dx   wi fdx wi EA( x )
0
dx
dx
dx
0
0
0
3
2
1
L
Finite Element Method – Piecewise
Approximation
u
x
u
x
FEM Formulation of Axially
Loaded Bar – Governing Equations
• Differential Equation
d 
du 
EA( x)   f ( x)  0

dx 
dx 
0 x L
• Weighted-Integral Formulation

L
0
d 

du 
w  EA( x)   f ( x) dx  0
dx 
 dx 

• Weak Form
L
 dw 

du 
du 

0     EA( x)   wf ( x) dx  w EA( x) 
dx 
dx 
dx  0


0 
L
Approximation Methods – Finite
Element Method
Example:
Step 1: Discretization
Step 2: Weak form of one element
P1
P2
x1
x2
x2
du 
du 
 dw 


EA
(
x
)

w
(
x
)
f
(
x
)
dx

w
(
x
)
EA
(
x
)


 0

x  dx 
dx
dx


 x1

1
x2
du 
 dw 

EA
(
x
)

w
(
x
)
f
(
x
)


dx  w x2 P2  w x1 P1  0
x  dx 
dx


1
x2
Approximation Methods – Finite
Element Method
Example (cont):
Step 3: Choosing shape functions
- linear shape functions
u u
1 1
x
x1
l
x2
 2u2
x1
x0
x1
1x
1 x
x2  x
x  x1


;


1
2
1 
; 2 
2
2
l
l
x  1l  x
2
x  x  x1   1; x 
1
l
2
x
Approximation Methods – Finite
Element Method
Example (cont):
Step 4: Forming element equation
E,A are constant
Let w  1 , weak form becomes
Let w  2 , weak form becomes
1
u u 
x  l  EA  2 l 1 dx  x 1 f dx  1P2  1P1  0
1
1
x2
x2
2
1
u2  u1 
EA

dx



l
 2 f dx  2 P2  2 P1  0
l 
x1 
x1
x2
x
 x2


fdx
 1

EA  1 1  u1   x1
  P1   f1   P1 

    




l  1 1  u2   x2
  P2   f 2   P2 
  2 fdx 
 x1

x
2
EA
EA
u1 
u2   1 f dx  P1
l
l
x1
x

2
EA
EA
u1 
u2   2 f dx  P2
l
l
x1
Approximation Methods – Finite
Element Method
Example (cont):
Step 5: Assembling to form system equation
Approach 1:
Element 1:
Element 2:
Element 3:
 1 1

E I A I  1 1
lI  0 0

0 0
0 0 u1I   f1I   P1I 
     
0 0 u2I   f 2I   P2I 
      
0 0  0   0   0 

0 0  0   0   0 
0 0 0

E II AII 0 1 1
l II 0 1 1

0 0 0
0

E III AIII 0
l III 0

0
0  0   0   0 
0 u1II   f1II   P1II 
      
0 u2II   f 2II   P2II 

0  0   0   0 
0  0   0   0 
0 0 0   0   0   0 
   


0 1 1 u1III   f1III   P1III 

0 1 1  u2III   f 2III   P2III 
0
0
Approximation Methods – Finite
Element Method
Example (cont):
Step 5: Assembling to form system equation
Assembled System:
 E I AI
 lI

I I
 E A

 lI

 0


 0

E I AI
 I
l
E I AI E II AII
 II
lI
l
E II AII
 II
l
0
0
E II AII
l II
E II AII E III AIII

l II
l III
E III AIII
 III
l




  P1I

f1I
  u1   f1   P1  
0
 u2   f 2   P2   f 2I  f1II   P2I  P1II 
  II
           II
III 
III 
E III AIII   u3   f 3   P3   f 2  f1   P2  P1 
 III
 u4   f 4   P4   f III   P III 
l

 

2
2
III III 
E A 

l III
0
Approximation Methods – Finite
Element Method
Example (cont):
Step 5: Assembling to form system equation
Approach 2: Element connectivity table
kije  KIJ
Element 1 Element 2 Element 3
1
1
2
3
2
2
3
4
local node
(i,j)
global node index
(I,J)
Approximation Methods – Finite
Element Method
Example (cont):
Step 6: Imposing boundary conditions and forming condense system
Condensed system:
 E I AI E II AII
 I  II
l
 l
II II
E A

  l II


0


E II AII
 II
l
II II
E A
E III AIII

l II
l III
E III AIII
 III
l


 u2   f 2   0 
III III
E A      
 III  u3    f 3    0 
l
 u4   f 4   P 
III III
E A 

l III

0
Approximation Methods – Finite
Element Method
Example (cont):
Step 7: solution
Step 8: post calculation
u  u11  u22

du
d
d
 u1 1  u2 2
dx
dx
dx
  E  Eu1
d1
d
 Eu 2 2
dx
dx
Summary - Major Steps in FEM
• Discretization
• Derivation of element equation
• weak form
• construct form of approximation solution
over one element
• derive finite element model
• Assembling – putting elements together
• Imposing boundary conditions
• Solving equations
• Postcomputation
Exercises – Linear Element
Example 1:
E = 100 GPa, A = 1 cm2
Linear Formulation for Bar Element
u
x
u1
u2
f(x)
P2
P1
L = x2-x1
x=x1
 P1   f1   K11
    
 P2   f 2   K12
K12   u1 
 

K 22  u 2 
 di d j
where K ij   EA
 dx dx
x1
x2
1
1
x=x1
x= x2

dx  K ji , f i 

2
x2
  f dx
i
x1
1
x=x2
x
Higher Order Formulation for Bar Element
x
u2
u1
u
u3
3
2
1
u(x)  u11 (x) u22 (x) u33 (x)
x
u2
u1
u
u3
3
2
1
u4
4
u(x)  u11 ( x )  u22 ( x )  u33 ( x )  u44 ( x )
u1
u
x
1
u2
2
u3
3
u4
4
……………
……………
un
n
u(x)  u11 ( x )  u22 ( x )  u33 ( x )  u44 ( x )        unn ( x )
Natural Coordinates and Interpolation Functions
x
x=-1
x
x=1
x=x1
x l
x 0
x= x2
x  x  x1
x1  x2
x
2
Natural (or Normal) Coordinate: x 
l/2
x
x=-1
x=1
1  
2
1
x
x=-1
x=1
1 
1
2
x x  1
3
2
x=-1
1
x 1
2
2
, 2 
x 1
2
,  2  x  1x  1,  3 
x  1x
2
x
x=1 1  
9 
1 
1
27
x  1 x  1 x  1
 x   x  x  1,  2 
16 
3 
3
16
3

3
3  
27
x  1 x  1 x  1, 4  9 x  1 x  1  x  1 
16
3
16
3 
3


4
Quadratic Formulation for Bar Element
 P1   f 1   K 11
    
 P2    f 2    K 12
P   f  K
 3   3   13
K 12
K 22
K 23
K 13   u1 
 

K 23  u2 
K 33  u3 
1
 di d j 
 di d j  2
where K ij   EA 
dx   EA 
d x  K ji


 dx dx 
 dx dx  l
x1
1
x2
and f i 
x2
 i f  dx 
x1
1
x=-1
1
l
2
 i f  dx , i, j  1, 2, 3
1
2
x=0
3
x=1
Quadratic Formulation for Bar Element
u1
f(x)
u2
P1
u( x )  u11 ( x )  u2 2 ( x )  u3 3 ( x )  u1
1 
x3
x=1
x x  1
2
 u2 x  1x  1  u3
,  2  x  1x  1,  3
2
x1  x2
x
2
x
l/2
d1 2 d1 2x  1


,
dx l d x
l
P3
x2
x=0
x1
x=-1
x x  1
u3
P2
l
d x  dx
2
d2 2 d2
4x

 ,
dx
l dx
l
x  1x
2

x  1x

2
dx 2

dx l
d3 2 d3 2x  1


dx l d x
l
Exercises – Quadratic Element
Example 2:
E = 100 GPa, A1 = 1 cm2; A1 = 2 cm2
Some Issues
Non-constant cross section:
Interior load point:
Mixed boundary condition:
k